1
$\begingroup$

This one follows on from the question I just asked about logarithms.. Turns out 1/x questions confuse me (sorry for bombarding your exchange with questions, this isn't homework or anything I am just trying to understand stuff)

Okay so I am trying to work out the phase of the following complex number 1/jw

My usual method would be to plot this on an Argand diagram and then take the inverse tangent to get the angle

So as an example if the number was 5+2jw this would be tan^-1(imag/real) = tan^-1(2/5)

However with 1/jw there is no real and the imaginary is 1/jw so I would attempt the same thing imag/real which is (1/jw)/0 =infinity.. Tan^-1(infinity) =90.. So why does Matlab tell me the answer is -90??

Ps I am very sorry for not using formatting I am on my ipad and also am not sure how to do it..

$\endgroup$

1 Answer 1

2
$\begingroup$

Take into acount that:

$$z = 1/j = j/j^2 = - j = 1|_{-\pi/2}.$$

If you actually want to obtain the phase of the complex number through the "tangent method", then:

$$\alpha = \arctan(\text{Im}(z)/\text{Re}(z)) = \arctan(-\infty) = - \pi/2.$$

Cheers!

$\endgroup$
3
  • $\begingroup$ Hi thank you for your response.. Please could you explain the pi/2 bit.. I understand that you have equated 1/j to j/j^2 and j^2 =-1 but then I get a bit lost.. $\endgroup$
    – Shasam
    Jan 16, 2014 at 11:47
  • $\begingroup$ Of course. When considering a complex number in cartesian form, say $z = a + b i$, it can be represented in polar form as follows: $z = M|_{\theta}$ where $M = \sqrt{a^2+b^2}$ is its modulus and $\theta = \arctan{b/a}$ is its argument. For example $1+i = \sqrt{2}|_{\pi/4}$. Cheers! $\endgroup$
    – Dmoreno
    Jan 16, 2014 at 11:52
  • $\begingroup$ Ahh right it was just a different kind of notation. I get it now.. There are too many ways to represent a complex number in my opinion! Thanks a Lot $\endgroup$
    – Shasam
    Jan 16, 2014 at 12:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.